Indeed a marvelous problem with such a simple condition, yet such a sophisticated and intriguing solution. Good luck to anyone who attempts it!

$f(\frac{\sqrt{2}}{2})=1010$

$f\text{ is even}$

Let f(x) = asin^(-1)(x) + b cos^(-1)(x) , a&b are real. Then discuss cases

Urvs wrote: Let $f(x) = asin^{-1}(x)+b cos^{-1}(x)$ , $a,b$ are real. Then discuss cases FIxed it for you

BearNo21 wrote: Urvs wrote: Let $f(x) = asin^{-1}(x)+b cos^{-1}(x)$ , $a,b$ are real. Then discuss cases FIxed it for you Thanks

Or u may use f(x) = Asin^(-1)(x) +B as sin^-1(x) +cos^-1(x) =pi /2

Urvs wrote: Or u may use $f(x) = Asin^{-1}(x)+B$ as $sin^{-1}(x)+cos^{-1}(x)=\frac{\pi}{2}$

Note that $$f(\sin(x))+f(\cos(x))=2020=f(\sin(-x))+f(\cos(-x)),$$so $f$ is even. Write $f(x)=g(1-2x^2)+1010$ with $g:[-1,1]\rightarrow \mathbb{R}$. The condition becomes $$g(\cos(2x))+g(-\cos(2x))=0$$for all $x\in \mathbb{R}$, which is equivalent to $g$ being odd. We can check that all $f$ of this form satisfy the given condition. Hence, the general solution is $$f(x)=g(1-2x^2)+1010$$for any odd function $g:[-1,1]\rightarrow \mathbb{R}$.

rmtf1111 wrote: Indeed a marvelous problem with such a simple condition, yet such a sophisticated and intriguing solution. Good luck to anyone who attempts it! Good sense of humor, man!

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